Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$A$ is symmetric positive definite and $A = LDL^T$, where $L$ is unit lower triangular and $D$ is diagonal. I want to prove that the main-diagonal entries of $D$ are all positive.

I have tried $\det(A)>0 \Rightarrow \det(LDL^T)>0$. Since $\det(L)=\det(L^T)=1$, $\det(D)$ must be positive. But, that doesn't mean all of the main diagonal entries of D are positive. Should I be using a different property?

share|cite|improve this question
If $X$ is non-singular and $A$ is PD then $X^T A X$ is PD. Take $X = L^{-1}$. – user2468 Feb 25 '12 at 20:16
@J.D. Do you mean let $X=(L^{-1})^T$? So that $X^TAX=D$ – Ashley Feb 27 '12 at 3:46
Yes. That's what I meant. – user2468 Feb 27 '12 at 4:12

There are probably many ways of doing it. Here is one:

Note first that $A$ admits a square root, i.e. there is a symmetric positive definite matrix $A^{1/2}$ such that $(A^{1/2})^2=A$. This is done by writing $A$, via the Spectral Theorem, as $$ A=\sum_{j=1}^n \lambda_j P_j, $$ where $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$ counting multiplicities, and $P_1,\ldots,P_n$ are pairwise orthogonal projections of rank one (i.e. the projections onto the corresponding eigenspaces). Then define $$ A^{1/2}=\sum_{j=1}^n \lambda_j^{1/2} P_j. $$

Now, since $L$ is invertible, we can write $$ D=MAM^T, $$ where $M=L^{-1}$. Considering the canonical basis $\{e_1,\ldots,e_n\}$, we have $$ D_{kk}=\langle De_k,e_k\rangle = \langle MAM^Te_k,e_k\rangle =\langle A^{1/2}M^Te_k, A^{1/2}M^Te_k\rangle\geq0. $$ But $D$ is invertible, so $D_{kk}\ne0$, and so $D_{kk}>0$ for all $k=1.\ldots,n$.

share|cite|improve this answer

Let $e_j$ the $j$-th vector of the canonical basis. Since $L$ is invertible we are allowed to write $(L^t)^{-1}e_j$ and since $A$ is positive definite $$0<((L^{-1})^te_j)^tA(L^{-1})^te_j=e_j^tL^{-1}A(L^{-1})^{t}e_j=e_j^tDe_j=D_{jj}$$ and we are done.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.